Can you figure this out and explain if possible?

Question

A car of weight 2720 N operating at a rate of 156 Kw develops a maximum speed of 31 m/s on a level, horizontal road.
Assuming that the resistive force (due to friction and air resistance) remains constant, what is the car's maximum speed on an incline of 1 in 20, i.e. if @ (theta) is the angle of the incline with the horizontal, sin @ = 1/20? Answer in units of m/s

A 172 kg crate is pulled along a level surface by an engine. The coefficient of kinetic friction between the crate and the surface is 0.496. The acceleration of gravity is 9.8 m/s2
How much power must the engine deliver to move the crate at a constant speed of 7.53 m/s? Answer in units of watts.

First correct answerer gets 10 points!

Answer 1

Power = force * velocity (when they're in the same or opposite directions, as they are in these examples).
1) On the level, the friction force is power/velocity = 156000/31 = 5032 N. On the incline there is an added (or subtracted; you didn't say whether it's up or downhill) force of .05w=136 N. The speed is power/force = 156000/(5032+136) = 30.19 m/s uphill, 156000/(5032-136) = 31.96 m/s downhill. BTW, this is not a realistic problem: a very lightweight car with a huge amount of friction & drag force.
2) Normal force = vertical force (since it's level) = mg = 1687 N. Friction force Ff = coefficient of friction * normal force = .496 * 1687 = 836 N. Power = Ff*v = 836 * 7.53 = 6300 w.

Answer 2

1. On a horizontal road, the frictional force is given by P = Fv => F = P/v = 156 x 10^3 / 31 = 5032 N.
On the incline, draw out the forces involved: gravitational (straight down), reaction from the road (normal to road), car engine (along road) and frictional (along road in opposite direction). Resolve the gravitational force into components along and normal to the road. The normal component will be balanced by the reaction force. The component along the road will be 2720 sin @ = 2720 / 20 = 136 N, and will be in the same direction as the frictional force if going up the incline, the other direction going down.

So the combined opposing force will be 5032+136 = 5168 N going up the incline, or 5032-136 = 4896 N going down. The maximum speed is given by P = Fv => v = P/F, and is thus 30 m/s going up, or 32 m/s going down.

2. The gravitational force is F = mg = 172(9.8) = 1686 N. The frictional force is therefore 1686 * 0.496 = 836 N. The required power is given by P = Fv = 836 (7.53) = 6.30 x 10^3 W.